Answer: (a) \(\alpha=-\frac{\pi}{4},\frac{3\pi}{4}\); (b)(ii) \(\phi=20^\circ,100^\circ,140^\circ\).

(a) Since

the equation becomes

So

\(\frac1{\sin\alpha}+\frac1{\cos\alpha}=0.\)

Multiplying by \(\sin\alpha\cos\alpha\),

\(\cos\alpha+\sin\alpha=0.\)

Hence

\(\tan\alpha=-1.\)

For \(-\pi\lt \alpha\lt \pi\),

\(\alpha=-\frac{\pi}{4},\quad \frac{3\pi}{4}.\)

(b)(i) Start with the left-hand side:

Use a common denominator:

Expand the numerator:

\(\cos^2\theta+1-2\sin\theta+\sin^2\theta.\)

Since \(\sin^2\theta+\cos^2\theta=1\), this becomes

\(2-2\sin\theta=2(1-\sin\theta).\)

Therefore

(b)(ii) Using the identity with \(\theta=3\phi\),

\(2\operatorname{sec}3\phi=4.\)

Thus

\(\operatorname{sec}3\phi=2,\)

so

\(\cos3\phi=\frac12.\)

Since \(0^\circ\leq\phi\leq180^\circ\),

\(0^\circ\leq3\phi\leq540^\circ.\)

In this interval,

\(3\phi=60^\circ,\ 300^\circ,\ 420^\circ.\)

Therefore

\(\phi=20^\circ,\ 100^\circ,\ 140^\circ.\)